Which of the following compound have number of `p pi-p pi` bond equal to `p pi - d pi` bond.

Options

A. `SO_(3)`
B. `SO_(2)`
C. `SOCl_(2)`
D. `SO_(2)Cl_(2)`

Answer

To determine which compound has an equal number of \( p\pi - p\pi \) bonds and \( p\pi - d\pi \) bonds, we will analyze the hybridization and bonding characteristics of the given compounds. Let's go through the analysis step-by-step.

### Step 1: Analyze Sulfur Trioxide (SO₃)
1. **Determine the hybridization**:
- Valence electrons of sulfur = 6
- Number of monovalent atoms (O) = 0
- Hybridization = \( \frac{1}{2} \times (6 + 0) = 3 \) → \( sp^2 \)

2. **Count the bonds**:
- Sulfur can form three \( p\pi - p\pi \) bonds with oxygen.
- Since all orbitals are used for bonding, there are no \( p\pi - d\pi \) bonds.

### Step 2: Analyze Sulfur Dioxide (SO₂)
1. **Determine the hybridization**:
- Valence electrons of sulfur = 6
- Number of monovalent atoms (O) = 2
- Hybridization = \( \frac{1}{2} \times (6 + 2) = 4 \) → \( sp^2 \)

2. **Count the bonds**:
- Sulfur forms two bonds with oxygen.
- There is one lone pair on sulfur, which means it forms:
- 1 \( p\pi - p\pi \) bond (with one oxygen)
- 1 \( p\pi - d\pi \) bond (with the other oxygen)

### Step 3: Analyze Sulfur Dichloride (SOCl₂)
1. **Determine the hybridization**:
- Valence electrons of sulfur = 6
- Number of monovalent atoms (Cl) = 2
- Hybridization = \( \frac{1}{2} \times (6 + 2) = 4 \) → \( sp^3 \)

2. **Count the bonds**:
- Since sulfur is \( sp^3 \), it does not have any \( p\pi \) orbitals available for bonding. Therefore, it cannot form any \( p\pi - p\pi \) or \( p\pi - d\pi \) bonds.

### Step 4: Analyze Sulfur Dichloride (SO₂Cl₂)
1. **Determine the hybridization**:
- Valence electrons of sulfur = 6
- Number of monovalent atoms (Cl) = 2
- Hybridization = \( \frac{1}{2} \times (6 + 2) = 4 \) → \( sp^3 \)

2. **Count the bonds**:
- Similar to SOCl₂, sulfur in SO₂Cl₂ is \( sp^3 \) and cannot form any \( p\pi \) bonds.

### Conclusion
From the analysis, we find that:
- **SO₃** has 3 \( p\pi - p\pi \) bonds and 0 \( p\pi - d\pi \) bonds.
- **SO₂** has 1 \( p\pi - p\pi \) bond and 1 \( p\pi - d\pi \) bond.
- **SOCl₂** has 0 \( p\pi - p\pi \) bonds and 0 \( p\pi - d\pi \) bonds.
- **SO₂Cl₂** has 0 \( p\pi - p\pi \) bonds and 0 \( p\pi - d\pi \) bonds.

Thus, the compound that has an equal number of \( p\pi - p\pi \) bonds and \( p\pi - d\pi \) bonds is **Sulfur Dioxide (SO₂)**.

### Final Answer
The answer is **Sulfur Dioxide (SO₂)**.